Xn+1=cube root 100+10Xn

ITERATIVE PROCESSES

find the value of x correct to 1 decimal place when X1=16

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1 Answer

After several iterations X becomes stable around 5.354866968.

x=∛(100+10x), x³=100+10x, x³-10x-100=0. The root of this equation is the value of X calculated through the iteration.

X₁=∛100+160=∛260=6.3825 approx.

X₂=∛100+63.825=∛163.825=5.4718 approx.

X₃=∛154.718=5.3684 approx.

X₄=∛153.684=5.3564 approx, so x=5.4 to 1 dec place.
by Top Rated User (1.2m points)

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